Question 732700
There will be 2 solutions for {{{ s = 240 }}}
You must solve for {{{ t }}} for both solutions
and then find {{{ t[2] - t[1] }}}
{{{ s = -16t^2 + 128t }}}
{{{ 240 = -16t^2 + 128t }}}
{{{ 15 = -t^2 + 8t }}}
{{{ -t^2 + 8t - 15 = 0 }}}
Use quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -1 }}}
{{{ b = 8 }}}
{{{ c = -15 }}}
{{{ t = (-8 +- sqrt( 8^2 - 4*(-1)*(-15) )) / (2*(-1)) }}}
{{{ t = (-8 +- sqrt( 64 - 60 )) / (-2) }}}
{{{ t = (-8 +- sqrt( 4 )) / (-2) }}}
{{{ t = (-8 + 2) / (-2) }}}
{{{ t = 3 }}}
and
{{{ t = (-8 - 2) / (-2) }}}
{{{ t = 5 }}}
{{{ 5 - 3 = 2 }}}
The interval of time the ball is more than 240 above the ground is 2 sec
Here's the plot:
{{{ graph( 400, 400, -1, 10, -30, 300, -16x^2 + 128x ) }}}